Asymptotics of Tracy-Widom distributions and the total integral of a Painlevé II function

Transcription

1 Asymptotics of Tracy-Widom distributions and the total integral of a Painlevé II function Jinho Baik, Robert Buckingham and Jeffery DiFranco April 6, 007 Abstract The Tracy-Widom distribution functions involve integrals of a Painlevé II function starting from positive infinity. In this paper, we express the Tracy-Widom distribution functions in terms of integrals starting from minus infinity. There are two consequences of these new representations. The first is the evaluation of the total integral of the Hastings-McLeod solution of the Painlevé II equation. The second is the evaluation of the constant term of the asymptotic expansions of the Tracy-Widom distribution functions as the distribution parameter approaches minus infinity. For the GUE Tracy-Widom distribution function, this gives an alternative proof of the recent work of Deift, Its, and Krasovsky. The constant terms for the GOE and GSE Tracy-Widom distribution functions are new. Introduction Let F x, F x, and F 4 x denote the GOE, GUE, and GSE Tracy-Widom distribution functions, respectively. They are defined as [4, 5 F x = FxEx, F x = Fx, F 4 x = { Ex + } Fx, Ex where Fx = exp x Rsds, Ex = exp Here the real function qx is the solution to the Painlevé II equation that satisfies the boundary condition Recall [ that the Airy function Aix satisfies Ai x = xaix and Aix x qsds. q = q 3 + xq, 3 qx Aix, x +. 4 πx /4 e 3 x 3/, x +. 5 Department of Mathematics, University of Michigan, Ann Arbor, MI, 4809, baik@umich.edu Courant Institute of Mathematical Sciences, New York University Department of Mathematics, University of Michigan, Ann Arbor, MI, 4809, robbiejb@umich.edu Department of Mathematics, University of Michigan, Ann Arbor, MI, 4809, jeffcd@umich.edu

2 There is a unique global solution qx to the equation 3 with the condition 4 the Hastings-McLeod solution [9. The function Rs is defined as Rx = x qs ds. 6 By taking derivatives and using 3 and 4 see, for example,.5 of [4 and.6 of [3, the function Rx can also be written as Rx = q x xqx qx 4. 7 Integrating by parts, Fx can be written as Fx = exp x s xqs ds, 8 which is commonly used in the literature. Notice that involves integrals from x to positive infinity. The main results of this paper are the following representations of Fx and Ex, which involve integrals from minus infinity to x. Theorem.. For x < 0, { Fx = /48 e ζ e 4 x 3 exp x /6 where ζz is the Riemann-zeta function, and x Ex = { x /4 e 3 x 3/ exp qy Ry 4 y + } dy, 9 8y } y dy. 0 Remark. The formula 9 also follows from the recent work [8 of Deift, Its, and Krasovsky. See subsection. below for further discussion. The integrals in 9 and 0 converge. Indeed, it is known that [9, x qx = + 8x x x 9 + O x, x. This asymptotic behavior of q was obtained using the integrable structure of the Painlevé II equation see, for example, [6. The coefficients of the higher terms in the above asymptotic expansion can also be computed recursively see for example, Theorem.8 of [. For Rx, 7 and imply that Rx = x 4 x x 6 8 6x 9 + Ox We now discuss two consequences of Theorem... Total integrals of qx and Rx Comparing with, Theorem. is equivalent to the following. Corollary.. For c < 0, and c Rydy + c c, x. Ry 4 y + dy = 8y 4 log ζ + c 3 + log c 3 8 c qydy + qy y dy = log + 3 c 3/. 4

3 These formulas should be compared with the evaluation of the total integral of the Airy function [: Aiydy =. 5 Recall that the Airy differential equation is the small amplitude it of the Painlevé II equation. Unlike the Airy function, Rx and qx do not decay as x, and hence we need to subtract out the diverging terms in order to make the integrals finite.. Asymptotics of Tracy-Widom distribution functions as x Using formulas and, Tracy and Widom computed that see Section.D of [4 as x, e x 3 F x = τ + 3 x /8 6 x 3 + O x 6 6 for some undetermined constant τ. The constant τ was conjectured in the same paper [4 to be τ = /4 e ζ. 7 This conjecture 7 was recently proved by Deift, Its, and Krasovsky [8. In this paper, we present an alternative proof of 7. Moreover, we also compute the similar constants τ and τ 4 for the GOE and GSE Tracy-Widom distribution functions. The asymptotics similar to 6 follow from and : as x, F x = τ e 4 x 3 x /6 F 4 x = τ 4 e 4 x 3 + x /6 3 x 3/ 3 x 3/ Using and, Theorem. implies the following. + 4 x + 3/ O x 3, 8 4 x 3/ + O x 3. 9 Corollary.3. As x, Fx = /48 e ζ e 4 x 3 x / x 3 + O x 6 0 and Hence Ex = /4 e 3 x 3/ 4 x + 3/ O x 3. τ = /48 e ζ, τ = /4 e ζ, τ 4 = 35/48 e ζ. Conversely, using and, this Corollary together with implies Corollary., and hence Theorem.. This is one example of so-called constant problems in random matrix theory. One can ask the same question of evaluating the constant term in the asymptotic expansion in other distribution functions such as the iting gap distribution in the bulk or in the hard edge. For the gap probability distribution in the bulk scaling it which is given by the Fredholm determinant of the sine-kernel, Dyson [3 first conjectured the constant term for β = in terms of ζ using a formula in an earlier work [7 of Widom. This conjecture was proved by Ehrhardt [4 and Krasovsky [0, independently and simultaneously. A third proof was given in [9. The constant problem for β = and β = 4 ensembles in the bulk scaling it was recently obtained by Ehrhardt [5. For the hard edge of the β-laguerre ensemble associated with the weight x m e x, the 3

4 constant was obtained by Forrester [8 equation.6a when m is a non-negative integer and /β is a positive integer. The above iting distribution functions in random matrix theory are expressed in terms of a Fredholm determinant or an integral involving a Painlevé function. For example, the proof of [8 used the Freldhom determinant formula of the GUE Tracy-Widom distribution: where A x is the operator on L x, whose kernel is F x = det A x, 3 Au, v = AiuAi v Ai uaiv. 4 u v In terms of the Fredholm determinant formula, the difficulty comes from the fact that even if we know all the eigenvalues λ j x of A x, we still need to evaluate the product j= λ jx. When one uses the Painlevé function, one faces a similar difficulty of evaluating the total integral of the Painlevé function. We remark that the asymptotics as x + of Fx and Ex and hence F β x are, using 4,.3 Outline of the proof Fx = e 4 3 x3/ 3πx 3/ Ex = e 3 x3/ 4 πx 3/ 35 4x + 3/ Ox x 3/ + Ox 3, 5. 6 The Tracy-Widom distribution functions are the its of a variety of objects such as the largest eigenvalue of certain ensembles of random matrices, the length of the longest increasing subsequence of a random permutation, the last passage time of a certain last passage percolation model, and the height of a certain random growth model see, for example, the survey [. Dyson [3 exploited this notion of universality to solve the constant problem for the sine-kernel determinant. Namely, among the many different quantities whose it is the sine-kernel determinant, he chose one for which the associated constant term is explicitly computable specifically, a certain Toeplitz determinant on an arc for which the constant term had been obtained by Widom [7, and then took the appropriate it while checking the it of the constant term. However, the rigorous proof of this idea was only obtained in the subsequent work of Ehrhardt [4 and Krasovsky [0. In order to apply this idea for F x, the key step is to choose the appropriate approximate ensemble. In the work of Deift, Its, and Krasovsky [8, the authors started with the Laguerre unitary ensemble and took the appropriate it while controlling the error terms. In this paper, we use the fact that F β x is a double-scaling it of a Toeplitz/Hankel determinant. Let D n t denote the n n Toeplitz determinant with symbol fe iθ = e tcosθ on the unit circle: π D n t = det e tcos θ e ij kθ dθ π = π n n! π [ π,π n e t P n j= cos θj 0 j,k n k<l n e iθ k e iθ l n dθ j. Note that some references e.g. [7 define D n as an n + n + determinant, whereas others e.g. [5 use our convention. In studying the asymptotics of the length of the longest increasing subsequence in random permutations, in [, the authors proved that when j= 7 n = [t + xt /3, 8 as t, e t D n t F x. 9 4

5 The idea of the proof of 9 in [ is as follows. The Toeplitz determinants are intimately related to orthogonal polynomials on the unit circle. Let p j z = κ j z j + be the orthonormal polynomial of degree t cos θ dθ j with respect to the weight e π π π : p j e iθ p k e iθ t cos θ dθ e π = δ jk for j, k If κ j > 0 then p j is unique. We denote by π j z; t = π j z the monic orthogonal polynomial: p k z = κ k π j z. Then see, for example, [3 the leading coefficient κ j = κ j t is given by D j t κ j t = D j+ t. 3 As the strong Szegö it theorem implies that D n t e t as n for fixed t, the left-hand-side of 9 can be written as D e t q t D n t = D q+ t = κ q t. 3 The basic result of [ is that q=n q=n κ Ry q t t, t, q = [t + /3 yt/3 33 for y in a compact subset of R. In [, the notations vx = Rx and ux = qx are used. Hence formally, as t with n = t + xt /3, log e t D n t = log κ q t t /3 q=n x log Ry dy t /3 The first step of this paper is to write, instead of 3, e t D n t = e t n q= D q t D q t = e t n κ q= q x Rydy = log F x t. Here D 0 t :=. Then formally, we expect that as t with n = t + xt /3, 35 converges to an integral from to x. For this to work, we need the asymptotics of κ q t for the whole range of q and t such that q t + xt /3 as t. It turns out it is more convenient to write, for an arbitrary fixed L, e t D n t = e t D L t n q=l+ We introduce another fixed large number M > 0 and write loge t D n = t + logd L }{{} exact part D q t D q t = e t D L t [t Mt /3 + q=l+ logκ q } {{ } Airy part + n κ q=l+ q [t+xt /3 q=[t Mt /3 Since L and M are arbitrary, we can compute the desired it by computing 36 t. logκ q } {{ } Painlevé part. 37 D L,M t loge t n t, n = [t + xt /

6 From 33, the Painlevé part converges to a finite integral of Ry from y = M to y = x as t. For the Airy part, we need the asymptotics of κ q t for L + q t Mt /3 as t for fixed L, M > 0. The paper [ obtains a weak one-sided bound of κ q t for ǫt q t Mt /3 as t, where ǫ > 0 is small but fixed. The technical part of this paper is to compute the leading asymptotics of κ q t in L + q t Mt /3 with proper control of the errors so that the Airy part converges. The advantage of introducing L is that we do not need small values of q, which simplifies the analysis. The calculation is carried out in Section 3. Finally, for the exact part, the asymptotics of D L t as t are straightforward using a steepest-descent method since the size of the determinant is fixed and only the weight varies. The it is given in terms of the Selberg integral for the L L Gaussian unitary ensemble, which is given by a product of Gamma functions, the Barnes G-function. The asymptotics of the Barnes G-function as L are related to the term ζ see 48 below. The computation is carried out in Section. Now we outline the proof of the formula 0 for Ex. In the study of symmetrized random permutations it was proven in [4, 5 that, in a similar double scaling it, certain other determinants converge to F x and F 4 x. But it was observed in [4, 5 that these determinants can be expressed in terms of κ q t and π q 0; t for the same orthonormal polynomials 30 above. Hence by using the same idea for D n t, we only need to keep track of π q 0; t in the asymptotic analysis of the orthogonal polynomials. See Section 5 below for more details. This paper is organized as follows. In Section, the asymptotics of the exact part of 37 are computed. We compute the asymptotics of the Airy part in Section 3. The proof of 9 for Fx in Theorem. is then given in Section 4. The proof of 0 for Ex in Theorem. is given in Section 5. While we were writing up this paper, Alexander Its told us that there is another way to compute the constant term for Ex using a formula in [5. This idea will be explored in a later publication together with Its to compute the total integrals of other Painlevé solutions, such as the Ablowitz-Segur solution. Acknowledgments. The authors would like to thank P. Deift and A. Its for useful communications. The work of the first author was supported in part by NSF Grant # DMS and the Sloan Fellowship. The second and third authors were partially supported by NSF Focused Research Group grant # DMS The exact part We compute the exact part of 37. From equation 7, D L t = π L e t P L j= cos θj L! [ π,π L k<l L e iθ k e iθ l L dθ j. 39 Following the standard stationary phase method of restricting each integral to a small interval ǫ θ ǫ and expanding e iθ and e tcos θ in Taylor series, D L t is approximately π L e tl t P L j= θ j L! [ ǫ,ǫ L k<l L as t. By extending the range of integration to R L, we obtain j= L θ k θ l dθ j 40 e tl D Lt t π L DHerm L t =, 4 j= where DL Herm t = e t P L j= θ j L! [, L k<l n L θ k θ l dθ j. 4 j= 6

7 This integral is known as a Selberg integral and is computed explicitly as see for example, [, equation DL Herm π L/ L π L/ t = q! = GL +, 43 LL / t L / LL / t L / q=0 where Gz denotes the Barnes G-function, or double gamma function. Some properties of the Barnes G-functions are see, for example, [6, 6 Therefore, log G Gz + = ΓzGz 44 G = G = G3 =. 45 = 4 log 4 log π ζ, log G = 4 log + 4 log π + 3 ζ, 46 Gn =!! n!, n =, 3, 4,. 47 log Gz + = z log z 3 4 z + z logπ log z + ζ + O z as z. 48 logd L {Lt L L L t logt + log L 3 } 4 L + ζ = The Airy part The main result of this section is Lemma 3.0 which computes the Airy part of 37. We use the notation [t Mt /3 logκ L,M t q, 50 q=l+ γ = t q. 5 It is well known that the leading coefficients κ q of orthonormal polynomials can be expressed in terms of the solution of a matrix Riemann-Hilbert Problem RHP [7. We start with the RHP for m 5 defined in Section 6 p.56 of [, which is obtained through a series of explicit transformations of the original RHP for orthogonal polynomials. For notational ease, we drop the tildes Let θ c be defined such that 0 < θ c < π and sin θc = γ. For q in the regime L + q t Mt/3, we have γ >. Define the contours C = {e iθ : θ c < θ π} and C = {e iθ : 0 θ θ c } with the orientations given as in Figure a. Also define the contours C in and C out as in Figure a. Let Σ 5 = C C C in C out. Now let m 5 z = m 5 z; t, q be the solution to the following RHP: m 5 is analytic in z C\Σ 5 m 5 + z = m5 zv5 z for z Σ 5 5 m 5 = I as z 7

8 where the jump matrix v 5 z = v 5 z; t, q is given by 0 0 v 5 e qα z = 0 0 e qα z C z C z C in C out. 53 Here αz = γ z s + 4 ξ s s ξs ξds, 54 where ξ = e iθc and the branch is chosen to be analytic in C\C and s ξs ξ +s for s. Then ξ O ξ ξ O ξ ξ C + C + in C + out 0 C C in C C out 0 C C in C C out 0 C C in C out ξ O ξ ξ O ξ ξ a The contour Σ 5 for m 5. b Introduction of O ξ and O ξ. c The contour Σ for m R. see 6.40 of [ and Figure : Contours used in the definition of the RHP for m 5 and m R. κ q = eq γ+log γ+ m π q 0 = q m We analyze the solution m 5 to this RHP for the regime L+ q t Mt /3 as t. Our analysis builds on the work of [ and makes two main technical improvements. The first is that the paper [ only considered the regime when ǫt q. Hence q necessarily grows to infinity. In this work, we allow q to be finite. The second is that we compute a higher order correction explicitly to the asymptotics obtained in [. This higher-order correction contributes to the sum 50. In [, only a one-sided bound of a similar sum was obtained. We merely outline the analysis for the parts that overlap with the analysis of [. From the construction of α in [ we have that e αz < for z C and e αz < for z C in C out. If we formally take the it of our jump matrix v 5 as q the jumps on the contours C in and C out approach the identity matrix and the jumps on C and C approach constant jumps. This iting RHP is solved explicitly by m 5, z = β + β i β β β β β + β, 57 i 8

9 where βz = z ξ /4, z ξ which is analytic for z C \ C and β as z. Note that m 5, 0 = γ = q t, γ t q m5, 0 = =. 58 γ t However, the convergence of the jump matrix v 5 z is not uniform near the points ξ and ξ, since αξ = αξ = 0. Therefore a parametrix is introduced around these points. For fixed δ < 00, define O ξ = {z : z ξ δ ξ ξ }, O ξ = {z : z ξ δ ξ ξ }. 59 Note that the diameter of O ξ is of order γ γ and varies as t and q vary. The diameter approaches 0 as γ or γ, which happens when q is close to t Mt /3 or L +, respectively. However, the point is that in the regime L + q t Mt /3, the diameter of O ξ cannot shrink too fast. Therefore, the usual Airy parametrix for a domain of fixed size still yields a good parametrix for the RHP in the regime under consideration. The case when γ slowly was analyzed in [ for the leading asymptotics of m 5. In this section, we also analyze the case when γ slowly, and also improve the work in [ to obtain a higher-order correction term. Orient the boundary of both O ξ and O ξ in the counterclockwise direction. Now as in [ see also [ for z O ξ \ Σ 5 define the matrix-valued function m p as πe m p z = i i iπ/6 q σ αz σ /3 z ξ Ψ z ξ qαz e qαzσ3, 60 where ω = e πi/3 and Ais Aiω s Ai s ω Ai ω s Ais Aiω s Ai s ω Ai ω s Ψs = Ais ω Aiωs Ai s Ai ωs Ais ω Aiωs Ai s Ai ωs e πi 6 σ3, 0 < args < π 3, e 0 πi 6 σ3 e πi 6 σ3 < args < π, 0, π < args < 4π 3,, π 3 e πi 6 σ3 4π, 3 < args < π. 6 We can define m p z = m p z for z O ξ by 60. For z / Σ 5 O ξ O ξ, let m p z = m 5, z. It is shown in [ that m p then solves a RHP that has the same jump conditions as m 5 on the contour C as well as on Σ 5 O, where we define O = O ξ O ξ. Define Rz = m 5 zmp z. Then R solves a RHP on Σ = O C C in C out O c with jump v R = m p v 5 vp m p. Explicitly, the jump matrix v R is given by 0, z Σ 5 O C, m 5, 0 e qα m 5,, z C O c, 0 0 v R = I + 6 m 5, 0 0 e qα m 5,, z C in C out O c, 0 R + ve R, z O, qα vqα 9

10 where v qα R is given explicitly in Lemma 3. below, and the matrix ve R is defined as ve R = v R I qα vqα R for z O. Since m 5 0 = R0m p 0 = R0m 5, 0, we have m 5 0 = γ R 0 γ R 0 and m 5 0 = γ R 0 γ R 0. Therefore κ q = γ e q γ+log γ+ R 0 γ R 0 63 and π q 0 = q γ R 0 γ R In [, for z O, the jump matrix v R is approximated by the identity matrix I and the terms qα vqα R + ve R are treated as an error for the case when et q. For our purpose, we need to compute the contribution from the next order term qα vqα R explicitly. It will be shown in the following subsections that for any ǫ > 0, there are L 0 and M 0 such that for fixed L L 0 and M M 0, there is t 0 = t 0 L, M such that v R I L Σ < ǫ for all t t 0 and L + q t Mt /3. This was shown in [ for ǫ t q t Mt /3. Then we proceed via the standard Riemann-Hilbert analysis as, for example, in [. Let Cfz := fs πi Σ s z ds be the Cauchy operator defined for z / Σ. For z Σ, C fz is defined as the nontangential it of Cfz as z approaches z from the right-hand side of Σ. Define the operator C R f = C fv R I for f L Σ and the function µ = I + C R C R I. A simple scaling argument shows that C R is a uniformly bounded operator for L + q t Mt /3. Since the supremum norm of v R I can be made as small as necessary, we find that for L and M fixed but chosen large enough, C R is a bounded L operator with norm uniformly bounded for t sufficiently large for all q such that L + q t Mt /3. By the theory of Riemann-Hilbert problems, Rz I = µsv R I ds. 65 πi s z is, Also define Σ Define the contours Σ ± as the part of Σ in the upper-half and lower-half planes, respectively. That Σ ± = Σ ±Iz > C ± = C Σ ±, C ± in = C in Σ ±, C ± out = C out Σ ± 67 as shown in figure c. Now, by the Schwartz-reflexivity of v R see [, p. 59 and µ, and therefore We write this as where πi µsv R s I Σ s [ R0 I = R πi ds = πi µsv R s I Σ s + µsv R I Σ s + ds 68 ds. 69 R0 I = R + R + R 3 + R 4 + R 5 70 [ R = R πi O ξ qα vqα R sds, s [ [ R = R µs v E πi Rs ds, R 3 = R O ξ s πi C + µsv R s I ds, 7 s 0

11 [ R 4 = R πi Hence using 63, C + in C+ out [ µsv R s I ds, R 5 = R µs I s πi O ξ qα vqα R sds. s [t Mt /3 = = q=l+ [t Mt /3 q=l+ [t Mt /3 q=l+ logκ q { q γ + log γ + + log γ log R 0 γ R 0 } { q t q + log t + + t q log q log + R 0 γ R 0 log + [ 3. Calculation of R = R πi O ξ qα vqα R sds s First, we compute v qα R explicitly. 5 i= R i 0 γ R i 0 + R 0 γ R 0 Lemma 3.. For z O ξ, the jump matrix v R z can be written as v R = I + qα vqα R + ve R where v qα R = d β + c β id β c β id β c β d β + c β, c = 5 7, d = 7 7 } and vr E = O qα. 74 Proof. On O ξ, v R = m p m p+. On this contour m p = m 5, and m p+ is given by 60. Thus for z O ξ 3 /3 3 σ 3 4 v R = m 5, e qαzσ3 Ψ qαz αz 3 z ξ e iπ/6 q σ 3 6. z ξ π i i 75 For 0 < args < π 3, consider see 6 Ais Aiω Ψs = s Ai s ω Ai ω e s iπ/6σ3 76 with ω = e iπ/3. From Abramowitz and Stegun [ and 0.4.6, for args < π, Ais = e /3s3/ πs /4 Ai s = s/4 e /3s3/ π wherein c = 5 7 and d = 7 7. Also note the identity c + O 3 s3/ s 3 d + O 3 s3/ s Ais + ωaiωs + ω Aiω s = 0 79

12 and, from Abramowitz and Stegun , W[Ais, Aiωs = π e iπ/6 80 W[Ais, Aiω s = π eiπ/6 8 W[Aiωs, Aiω s = π eiπ/. 8 Using det Ψs = W[Ais, Aiω s = π eiπ/6, we have ω Ψ s = π Ai ω s Aiω s Ai se iπ/3 Aise iπ/3. 83 Using 3 λz3/ = αz, equations 77 and 78 yield Aiq /3 λz = e qα c πq /3 λ /4 qα + O qα Ai q /3 λz = q/3 λ /4 e qα d π qα + O qα Aiω q /3 e iπ/6 e qα λz = + c πq /3 λ /4 qα + O qα ω Ai ω q /3 λz = ω q /3 λ /4 e qα πe iπ/6 + d qα + O qα We insert the asymptotics into 83 resulting in the asymptotic formulas for 0 < argq /3 λz < π and qα large, Ψ q /3 λz = { πe iπ/6 e qασ3 i i + } σ3/6 d c 3 + O qα id ic qα qα. 88 It is straightforward to compute an analogous expansion for Ψ s for the other values of argq /3 λz in equation 76. To do this one must use the asymptotic formulas 77 and 78 as well as the additional expansions and from Abramowitz and Stegun [. Namely, for args < π/3, Ai s = s /4 [sin π 3 s3/ + π 4 Ai s = s/4 [cos π 3 s3/ + π 4 3c cos s3/ 3 s3/ + π + O 4 s 3, 89 3d sin s3/ 3 s3/ + π + O 4 s After carrying out this computation, the first two terms in the expansion are the same in all four regions. In other words, 88 is valid not only for 0 < argq /3 λz < π/3 but for all regions in the definition of Ψ in 76. Inserting the expansion in 88, equation 75 reduces to for all z O ξ. v R = I + qα Now we explicitly evaluate R. d β + c β id β c β id β c β d β + c β + O qα, 9

13 Lemma 3.. We have R = 8qγ 4qγ γ / 8qγ / 4qγ 8qγ / γ / 4qγ 8qγ + 4qγ. 9 Proof. From Lemma 3., it is sufficient to compute the integrals I = βs πi Oξ αss ds and I = πi O ξ βs ds. 93 αss We will use the relations ξ = e iθc and γ / sinθ c =, cosθ c = γ γ γ, sin θc = γ, / cos θc = / γ. 94 γ Note that αz = 3 z ξ3/ Gz for an analytic function Gz in O ξ see the bottom line at p.57 of [. Hence by residue calculations, I = πi z ξz ξ / Gzz dz = 3 95 ξ ξ / Gξξ and O ξ 3 I = 3z ξ πi Oξ / z ξ Gzz dz = 3 [ d z ξ / dz Gzz = 3 [ ξ ξ / Gξξ ξ ξ/ G ξ ξ ξ/ Gξ ξ Gξξ. But since αz = 3 z ξ3/ Gz and α z = γ 4 that and Using 94, we obtain and Therefore, and Gξ = z ξ z=ξ 96 z+ z z ξz ξ, a straightforward computation yields α z z ξ = γ ξ + / 4 ξ ξ ξ /. 97 G ξ = 3γ ξ + ξ ξ / 0 ξ 3 I = I = 3 4γ 3 5γ + ξ + ξ ξ ξ / γ + 3 i, 4γ / γ / 4γ / 5γ R = R = [ q R 7 7 I I = 8qγ 4qγ i R = R = [ 7 q I 7 I I γ / =. 0 8qγ / 4qγ 3

14 [ 3. Bound on R = R πi O ξ µs O qα We begin by establishing the leading term of αz for z near ξ. Lemma 3.3. For q < t and for z such that z ξ min{, ξ ξ }, αz 3 γ 3/4 z ξ 3/ e 3iπ/4 e 3iθc/ 50γ 3/4 z ξ 5/. 03 ξ ξ Proof. Write z = ξ + ǫ. Then ǫ = z ξ min{, ξ ξ }. Under the change of variables s = ξ + ǫu, equation 54 for αz becomes Using ξ = e iθc and 94, we have αzǫ = γǫ3/ + ξ ξ ξ 4 ξ 0 ds s u + ξǫ / ξ ξ u + ξǫ +ξ u du ǫu γǫ3/ + ξ ξ ξ 4 ξ = γ 3/4 e 3iπ/4 ǫ 3/ = γ 3/4 z ξ 3/ e 3iπ/4 e 3iθc/. 05 For the integrand in 04, using the inequalities +w / w for w and +w 0 w for w, and using the fact that +ξ and, we obtain ξ ξ ξ ξ Therefore, we obtain ξǫ / ξ ξ u + ξǫ +ξ u + ǫu 50 ǫ ξ ξ Lemma 3.4. For L + q t Mt /3, there is a constant c > 0 such that R 50 z ξ =. 06 ξ ξ ct. 07 q 3/ t q 5/ Proof. On O ξ, z ξ = δ ξ ξ 40 ξ ξ. Hence from Lemma 3.3, we have αz 6 γ 3/4 z ξ 3/ = δ3/ 6 γ 3/4 ξ ξ 3/ = 4 3/ γ 3 δ3/ 08 γ for z O ξ. Therefore, as µ and s are bounded on O ξ, R c for some constants c, c > 0, as ξ ξ = 4γ / γ. [ 3.3 Bound on R 3 = R πi C + O ξ γ 3 q γ 3 ds = πc γ 3 δ ξ ξ q γ 3 = µsv R s I ds s cγ q γ 5/ 09 Since µs and /s are bounded on C +, R3 c v R I L C + for some constant c > 0. But on C +, v R z I = Oe qαz. Hence R 3 c e qαz L C +, 0 4

15 for some constant c > 0. For z = e iθ C + hence θ c < θ π, using e iφ ξe iφ ξ = e iφ e iφc e iφ e iφc / e iπ+φ/, we have αe iθ = γ 4 = γ θ θ + e iφ θ c e iφ φ θ c cos e iφ e iθc e iφ e iθc ie iφ dφ sin / φ + θc φ sin / θc dφ. Recall that γ = t q. Note that αξ = 0, αs is real and positive on C+, and αeiθ increases as θ increases. Lemma 3.5. For q t, αe iθ γγ θ θc 3 π for θ c θ π. Proof. We consider two cases separately: θ c π 3 and θ c π 3. Start with the case when θ c π 3. We consider two sub-cases: θ π 3 and θ π 3. When θ π 3, 0 φ π 3, 0 φ+θc π φ θc and 0 π 3. Hence using the basic inequalities cosx for 0 x π 3 and sinx π x for 0 x π, we find that αe iθ γ π θ θ c φ + θ c / φ θ c / dφ γ π When θ π 3, from the monotonicity of αeiθ and using 4, π θ αe iθ αe π 3 i γ 4π θ π 3 θ c θ c φ θ c dφ = γ 4π θ θ c For 0 θ c π 3 and π 3 θ π, we have π 3 θ c π 3 3 θ θ c. Therefore, αe iθ γ. θ θc 6 π For the second case, when θ c π 3, using the change of variables φ π φ, π θc φ φ + π αe iθ = γ sin sin / θc π sin / θc φ dφ. 7 Note that 0 φ π 3, 0 φ+π θc π θ c π π θc φ 3, and 0 π 3. Using the basic inequalities sinx 3 3 π x for 0 x π 3 and sinx 3 3 4π x for 0 x π 3, we find that Since γ γ αe iθ 7γ 8 π π θc π θ π θc φφ + π θ c / π θ c φ / dφ 7γ 8 φπ θ c φdφ π π θ = 9γ 6 π θc + π θ θ θ c 9γ 6 π π θ cθ θ c. = cos θ c = sin π θc π θ c, we have αe iθ Combining 4, 6, and 9 completes the proof π γγ θ θc. 9 5

16 Lemma 3.6. For L + q t Mt /3, there is a constant c > 0 such that R 3 ct. 0 q 3/ t q 5/ Proof. Let e iθ be the endpoint of C + on O ξ. Note that since radius of O ξ is δ ξ ξ = 4δ γ Using Lemma 3.5 and changing variables, where e qα L C + π θ e Using the inequality a e x dx a 3 for a > 0, γ, γ θ θ c 4δ. γ q γγ 6π θ θ c dθ 3π q γγ / x e x dx / q γγ x = θ θ c 4δ 3/4 γ q. 3 3π 3π γ e qα L C + Hence from 0 we obtain 0. 3π q γγ [ 3.4 Bound on R 4 = R πi C + in C+ out / x 3 µsv R s I ds s 9π t 8δ 3. 4 q 3/ t q 5/ As before, since on C + in C+ out the functions µs and s are uniformly bounded and v Rz I = Oe qαz, for some constants c, c > 0. R 4 c v R I L C + in C+ out c eqα L C + in C+ out 5 Lemma 3.7. For L + q t Mt /3, there is a constant c > 0 such that R 4 ct. 6 q 3/ t q 5/ Proof. Let γ 0 = csc π 4. Let δ 4 > 0 be a small positive number defined on p. 5 of [. We estimate e qα in the following three cases separately: i t+δ 4 q t Mt /3 +, ii t γ 0 q t+δ 4 and iii L + q t γ 0. i For t + δ 4 q t Mt /3, from 6.37 of [, there are constants c, c, c 3, c 4, c 5 > 0 such that e qα L C + in C+ out c Using the change of variables y = c 3 qx 3, e qα L C + in C+ out c 3c /3 e c3qx 3 c γ 3 q /3 c dx + c 4 e c5q. 7 c qγ 3/ e y y /3 dy + c 4e c5q 3c c 3qγ e c3 c3qγ 3/ + c 4 e c5q c 3c 5 c 3 q γ + c 4 5/ c. 5 q 6 8

17 Since γ and γ + δ 4, we find that e qα L C + in C+ out cγ q γ 5/ = ct q 3/ t q 5/ 9 for a constant c > 0. ii For t γ 0 q t+δ 4, note that the radius of O ξ is of O. Hence a standard calculation in Riemann-Hilbert steepest-descent analysis shows that for z C + in C+ out, Rαz c for some constant c > 0. Since the length of L C + in C+ out is bounded, e qα L C + in C+ out c e c q c c q c γ q γ 5/ = c t q 3/ t q 5/ 30 for some constants c, c > 0. iii Consider the case when L + q t γ 0. Then 0 θ c π. In this case, we make a specific choice of C + in and C+ out: } C + in {ξ = + ρ sinθ c e iθc+ 7 6 π : δ ρ sinθ c π } 3 C out {ξ + = + ρ sin c e iθc 6 π : δ ρ sinθ c 6 π. The contours C + in are straight line segments from ξ to a point on the positive real axis. Recall that O ξ has the radius δ ξ ξ = δ sin θ c. Now we estimate Rαz for z C + in C+ out. For z C+ in, take the contour in 54 to be the straight line from ξ to z. Then one can check from the geometry that 0 arg + s θ c, 0 args θ c, args ξ = θ c + 7π 6 π args ξ 5π 6 θ c, argds = θ c + π 3 6. Therefore the argument of the integrand in 54 is in [π, π+ π 6 +θ c [π, π+ π 3 since 0 θ c π. Thus, the cosine of the argument is greater than or equal to cos π 3 =. Therefore, for z = ξ + ρ sin θ ce iθc+ 7 6 π C + in, by the change of variables s = ξ + y sin θ ce iθc+ 7 6 π, Rαz γ z + s 8 s s ξs ξ ds ξ γ sin θ c cos θc ρ 0 y + y sin θc cos θ ei θc + 7π 6 i + y sinθ c e i7π 6 yeiθc+ 7π 6 / dy. Using the inequality + xe iφ sin φ for all x R, and using 0 θ c π and y sinθ c+, we 7π 6 have ρ 33 Rαz γ sin θ c cos θc 36 ydy 0 3/ γ = ρ 3/ 34 7 γ 3/ γ δ 3/ 7 γ 7

18 for z C + in. For z C+ out, taking the contour in 54 to be the straight line from ξ to z, we can check that 0 arg + s θ c, 0 args θ c, args ξ = θ c π 6 π 6 θ c args ξ π, argds = θ c π Hence the argument of the integrand in 54 is in [ θ c π 6, θ c [ π 4, π 6. Therefore, for z = ξ + ρ sin θ c e iθc 6 π C out +, Rαz γ z 4 + s s s ξs ξ ds ξ ρ γ sin θ c cos θc 0 3/ γ 7 ρ 3/ γ 7 y + 3/ γ δ 3/. From 34 and 36, arguing as in 30, we obtain for a constant c > 0. Hence we obtain the estimate for R 4. [ 3.5 Bound on R 5 = R πi O ξ µs I γ y sin θc cos θ ei θc π 6 + y sin θ c e i π 6 i yeiθc π 6 / dy 36 e qα L C + in C+ out c t q 3/ t q 5/ 37 qα vqα R ds s Lemma 3.8. For L + q t Mt /3, there is a constant c > 0 such that R 5 ct. 38 q 3/ t q 5/ Proof. As µ I = C R C R I, and as C and C are uniformly bounded, µ I L Σ c 0 C R I L Σ = c 0 C v R I L Σ c v R I L Σ 39 for some constants c 0, c > 0 when t is large enough. Below, we assume that t is large enough so that the above estimate holds. Now R 5 R µs Ivqα ds O ξ qα s µ I L v qα R O ξ qα 40 L O ξ c v R I L v qα R Σ qα. Since βz = z ξ z ξ L O ξ /4 is bounded above and below for z Oξ, v qα R z in Lemma 3. is bounded. Using 08 and the fact that the radius of O ξ is δ ξ ξ = 4δ γ γ v qα R qα L O ξ, we have c γ q γ 5/ = c t q 3/ t q 5/ 4 8

19 for a constant c > 0. Write v R I L Σ = v R I L O + v R I L Σ\ O. As v R I = vqα R qα + ve R bounded by c3 qα for a constant c 3 > 0, we have, as in 4, for a constant c 4 > 0. On the other hand, v R I L O = v R I L O ξ in Lemma 3. is c 4 t q 3/ t q 5/ 4 v R I L Σ\ O = v R I L C + + v R I L C + in C+ out c 3 e qα L C + + eqα L C + in C+ out c 3 e qα L C + + eqα L C + in C+ out 43 for a constant c 3 > 0, since e qα for z C + and eqα for z C + in C+ out. Hence from 4 and 30, we have v R I L Σ\ O c 4 t q 3/ t q 5/ 44 for a constant c 4 > 0. By combining 4, 4, and 44, we obtain The Airy part From Lemmas 3.4, 3.6, 3.7, and 3.8, we find that, for L + q [t Mt /3, there is a constant c > 0 such that 5 R i 0 γ R i 0 ct q 3/ t q + ct 5/ q t q. 45 i= We need the following result. Lemma 3.9. We have and [t Mt /3 sup t = 0 46 M t q 3/ t q 5/ q=l+ [t Mt /3 sup t L t q = t q q=l+ Proof. We use the following basic inequality. Let a, b be integers. Let sx be a positive differentiable function in an interval [a, b + and there is c a, b such that s x < 0 for x [a, c and s x > 0 for x c, b +. Then b b+ sq sxdx. 48 As a function of 0 < q < t, q=a a t q 3/ t q 5/ decreases for 0 < q < 3 4 t and then increases for 3 4 t < q < t. 9

20 Hence [t Mt /3 sup t M t q 3/ t q 5/ q=l+ sup M t = M sup t t Mt /3 L t q 3/ t q [ 4 3t + 6tq q 3tt q 3/ q / 5/ dq t Mt /3 L = As a function of 0 < q < t, t q t q decreases for 0 < q < t and then increases for t < q < t. Hence [t Mt /3 sup t L t q t q q=l+ L t = L t t Mt /3 L t q t q dq [ t + q qt q t log t q q t Mt /3 L = Now we prove the main result of Section 3. Lemma 3.0 Airy part. We have L,M t [t Mt /3 q=l+ logκ q t tl + L logt L log L L M3 8 log M + 4 log = 0. 5 Proof. We first prove that { [t Mt /3 [t Mt logκ L,M t q q=l+ + 8t q + } = 0. q /3 q=l+ q t q + log t + + q log t q 5 Using + R 0 γ R 0 = 8t q q and using Lemmas 3.4, 3.6, 3.7, and 3.8, we find that 5 i= Ri 0 γ R i 0 + R 0 γ R

21 for L + q [t Mt /3, when we take t large enough. Hence using log + x x for x, L,M t L,M [t Mt /3 q=l+ [t Mt 4 /3 t q=l+ using 45 and Lemma 3.9. On the other hand, from Lemma 3., [t Mt /3 q=l+ 5 i= log + Ri 0 γ R i + R 0 γ R 0 5 i= 0 i R 0 γ R i 0 [t Mt 4c /3 t L,M t q 3/ t q + t 5/ q t q = 0 q=l+ log + R 0 γ R 0 = [t Mt /3 q=l+ 54 log 8t q. 55 q Using x log x + x 0 for 0 x, [t Mt /3 q=l+ [t Mt /3 q=l+ t Mt /3 L log 8t q + q [t Mt /3 q=l+ 64t q + 48t qq + 44q 64t q + 96t t q + + q 44q dq 0 8t q + q 56 as t by evaluating the integral explicitly. Using 54 and 56 in 7, we obtain 5. Now we compute each term of the sum in 5. Note that [t Mt /3 = t Mt /3 ǫ for 0 ǫ <. We have [t Mt /3 and q=l+ q t [t Mt /3 q + = [t Mt /3 q=l+ Since for positive integer m m k= q=l+ t q = t L t L Mt/3 + ǫmt /3 + + ǫ 57 q + logt = t Mt/3 ǫ L logt 58 m k log k = m logm! k= logk! = m logγm log Gm +, 59

22 we find that [t Mt /3 q=l+ q log q = t Mt /3 ǫ log Γt Mt /3 ǫ log Gt Mt /3 ǫ + L + log ΓL + + log GL + Note that from Stirling s formula for the Gamma function and the asymptotcs 48 for the Barnes G-function, as z, z + z log Γz + log Gz + = log z z z 4 logπ + ζ + o. 6 Now using the fact that where γ is Euler s constant, we obtain t and t K [t Mt /3 q=l+ [t Mt /3 q=l+ q= 60 K q log K γ = 0, 6 8t q 8 logt + 8 logmt/3 = 0 63 q logt γ + Therefore, using 57, 58, 60, 63, 64, and 6, we obtain t [t Mt /3 q=l+ L = q q= q t q + log t + + q log t + q 8t q + q t tl + L logt = M3 8 log M + 4 log 4 logπ + L + L + ζ log ΓL + + log GL + + γ L q=. q 65 Hence using 6 and 6 again, we obtain 5. 4 Proof of 9 in Theorem.: computation of Fx Recall equation 37 which we rewrite here: loge t D n = t + logd L }{{} exact part [t Mt /3 + q=l+ logκ q } {{ } Airy part + [t+xt /3 q=[t Mt /3 logκ q } {{ } Painlevé part. 66

23 For the Painlevé part, from [, [t+xt /3 t q=[t Mt /3 logκ q = x M x = M By combining 67, 5, and 49, we obtain M Rydy Ry 4 y + dy + 8y x3 log x 8 + M3 + log M. 8 M,L t loge t D n t x = Ry 4 y + dy + 8y x3 log x log + ζ Proof of 0 in Theorem.: computation of Ex Let Set see [4 It is shown in Corollary 7. of [5 that for where x lies in a compact subset of R, we have I j t = π e tcos θ e ijθ dθ. 69 π π D ++ l t = deti j k t + I j+k+ t 0 j,k l, 70 D + l t = deti j k t + I j+k+ t 0 j,k l. 7 l = [t + x t/3, 7 / D ++ t e t l t FxEx = 0, 73 / t D + t e t l t FxEx = In [5, the above results are shown with x in place of x for the alternate scaling t = l bx l/3. With the above scaling l = [t+ x t/3, we find x = x+ x t /3 /3. Since x is in a compact set, and Ex and Fx are continuous, the above results follow. From equations 73 and 74, for a fixed x R, Fx Ex = t e t t D ++ l D + l, l = [t + x t/3. 75 Let π j z; t be the monic orthogonal polynomial of degree k with respect to the weight π etcos θ dθ on the unit circle, as introduced in Section. It is shown in Corollary.7 of [4 that cf. 3 above e t / D ++ l t = e t / t D + l t = κ j+ t π j+0; t = j=l j=l j=l κ j+ t + π j+0; t = j=l κ j+ + π j+ 0; t, 76 κ j π j+ 0; t, 77 3

24 where the last equalities in 76 and 77 use the basic identity see e.g [3 Using equations 76 and 77, we can write D ++ l = D++ L l j=l πk0; t = κ k κ. 78 k D ++ j D ++ j = D ++ l L j=l κ j t + π j0; t, 79 and hence we have D + l = D + L l j=l+ D + j D + j = D + L l j=l+ κ j t π j 0; t, 80 D ++ l D + l = D ++ L D + L l k=l l κ k t j=l [ + π j 0; t π j+ 0; t. 8 Using recall that D l = deti j k t 0 j,k l is the l l Toeplitz determinant given in 7 we find that D ++ l D + l = D ++ a D a = κ j t, 8 D b j=b l D l L D + L D L j=l Inserting this equation into 75, and using 9, Hence we find cf. 36 [ + π j 0; t π j+ 0; t. 83 Fx Ex t D++ L = F x e D + l L [ + π j 0; t π j+ 0; t. 84 t D L Ex t D++ L = e D + l L [ + π j 0; t π j+ 0; t, l = t D L j=l j=l [ t + x t/3. 85 In analogy to equation 37, we write log Ex = L,M t t + log D++ L D + L D L }{{} Exact part + [t+ x t/3 + j=[t M t/3 + [t M t/3 j=l log[ + π j 0; t π j+ 0; t } {{ } Airy part log[ + π j 0; t π j+ 0; t } {{ } Painlevé part. 86 We compute each term as in the case of log Fx. 4

25 Lemma 5.. The Painlevé part We have for x < 0 and M > 0, [t+ x t/3 t j=[t M t/3 + x = M qy log[ + π j 0; t π j+ 0; t y x 3/ dy + 3 M3/ Proof. This follows from the following results in Corollaries 7. of [5: for x in a compact subset of R, t log + π j+ 0; t + log Ex = 0, 88 t j=[t+ x t/3 j=[t+ x t/3 log π j+ 0; t + log Ex = We remark that in [5, the notations vx = Rx and ux = qx are used. Lemma 5.. The Airy part [t M t/3 L,M t log[ + π j 0; t π j+ 0; t t Proof. Using 64, [t M t/3 j=l j=l log[ + π j 0 π j+ 0 = Using 70 and the fact that γ R [t M t/3 + q=l log + [t M t/3 + q=l 3 M3/ L log = γ log + γ R 0; q R 0; q. 9 γ 0 R 0 = 0, which follows from Lemma 3., this equals q=l [t γ M t/3 + 5 i= + log + γ R i 0; q Ri 0; q. 9 γ γ + γ From Lemmas 3.4, 3.6, 3.7, and 3.8, the same estimate as in 45 holds for 5 i= γ R i 0; q Ri 0; q for L + q [t Mt /3. Therefore the same argument as in 54 implies that the second sum in 9 vanishes in the it. Therefore, [t M t/3 [t M t/3 + t q L,M t log[ + π j 0; t π j+ 0; t log + = t j=l We use the Euler-Maclaurin summation formula b fk = k=a Set fq = log b a fxdx + fa + fb + f b b a 6 4! q=l + Err, Err π t q + t and write [t M t/3 = t M t/3 ǫ where 0 ǫ <. Then b a f 3 x dx. 94 ft Mt /3 ǫ + + fl = log, 95 t 5

26 and Also using f 3 q 0, f t Mt /3 ǫ + + f L = t 6 4! Err π t Mt /3 ǫ+ L as t. Finally, changing variables to w = + t Mt /3 ǫ+ L Hence the result follows. Lemma 5.3. The exact part We have [ f 3 q dq = π f t Mt /3 ǫ + f L 0, 97 t q t L Mt t and setting δ = + t and δ = /3 +ǫ t, t q +δ log + dq = 4t wlogwdw t δ = 4t [w w log w + +δ 4 w w L t = t log D++ L D + L L log D L 3 M3/ L log + O. t /3 δ 98 = Proof. Note see [4 that [ D ++ L = det π [ = det iπ [ = det π π π π π π 0 [e ij kθ e ij+k+θ e tcos θ dθ 0 j,k L e ij+θ sink + θe t cos θ dθ 0 j,k L sinj + θ sin θ sink + θ sin θ sin θe tcos θ dθ. 0 j,k L 00 Changing variables to x = cosθ and noting that sinj+θ sin θ is a polynomial in x of the form j x j + a constant multiple of the Chebyshev polynomial of the second kind, we have D ++ L [ j+k+ = det π = L L L π L! A steepest descent analysis yields that t D++ L e tl t L +L / π L L! x j+k x e tx dx 0 j,k L [, L k<l L [0, L k<l L L x k x l y k y l j= L j= x j etxj dx j. 0 yj e yj dy j =. 0 6

27 The multiple integral is another Selberg integral corresponding to a Laguerre ensemble which is evaluated explicitly as see of [ Hence Similarly, [0, L k<l L [ D + L = det π [ = det y k y l L j= t D++ L e tl t L +L / π GL + GL L G 3 π [ = det π = det [ π π π π π π π π π e ij kθ + e ij+k+θ e tcos θ dθ e ij+ θ cos Setting x = cosθ and noting that cosj+ θ cos θ form j x j +, we have D + L = LL π L det yj e yj dy j = L! GL + GL G k + θ e tcos θ dθ 0 j,k L 0 j,k L cos j + θ cos k + θ e t cos θ dθ cos j + θ cos k + θ θ cos θ cos θ cos = LL π L L! A steepest-descent analysis implies that t D + L e tl t L L/ π L L! e tcos θ dθ = j,k L 0 j,k L. 05 is a constant multiple of a Jacobi polynomial P, j x of the [ + x x j+k x etx dx [, L k<l L [0, L k<l L x k x l L j= y k y l L j= 0 j,k L + x j x j e txj dx j. 06 y / j e yj dy j =. 07 The multiple integral is also a Selberg integral see of [, and we obtain t D + L e tl t L L/ π GL + GL + L G =. 08 Using 04, 08, and D L t e 4tL t GL =, 09 t L L+ π L which follows from Section the exact part for F x, we obtain [ log D++ L D + L L L log [GL + GLGL + t D L G G3 GL = 0. 0 π L The result is now proved using the properties 46 and 48 for the Barnes G-function. 7

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